I have read the following: http://www.feynmanlectures.caltech.edu/I_20.html#Ch20-S1

The formula for $\tau_{xy}$ is derived in this chapter: http://www.feynmanlectures.caltech.edu/I_18.html#Ch18-S2. In this derivation until equation 18.11 ($\Delta W=(xF_y-yF_x)\Delta\theta$) the term $xF_y-yF_x$ does not seem to be arbitrary. It would not make any sense the other way around like: $yF_x-xF_y$, because of the equations 18.6 ($\Delta x=-PQ\sin\theta=-r\,\Delta\theta\cdot(y/r)=-y\,\Delta\theta$) and 18.7 ($\Delta y=+x\,\Delta\theta$).

Then $\tau_{yz}$ and $\tau_{zx}$ are derived (in the first link) by symmetry. $$\begin{alignedat}{6} &\tau_{xy}~&&=x&&F_y&&-y&&F_x&&,\\[.5ex] &\tau_{yz}~&&=y&&F_z&&-z&&F_y&&,\\[.5ex] &\tau_{zx}~&&=z&&F_x&&-x&&F_z&&. \end{alignedat}$$

I have some ideas and wonder whether or not they are true: I have got the feeling that they are wrong but I have no idea why they should be wrong.

  1. So, since the formula for the torque in the $xy$-plane ($\tau_{xy}$) is not arbitrarily-derived, there is to torque $\tau_{yx}$, since $yF_x-xF_y$ would not be correct in equation 18.11.

  2. In my first link, below equation 20.1 there are 2 pictures which shall demonstrate how the letters $x$, $y$ and $z$ can be interchanged. How would that make sense? Since equations 18.6 and 18.7 are true, we "live" in a right-handed coordinate system and $\tau_{yx}$ would not make any sense.

  3. Since $\tau_{yx}$ would not make any sense, the arguments in my first link down to equation 20.9 show that the right-hand rule is not arbitrary.

What is wrong with my arguments? What does $\tau_{yx}$ mean?

  • 1
    $\begingroup$ Who does torques by component. The definition is $\vec{\tau} = \vec{r} \times \vec{F}$ in vector form. $\endgroup$ Aug 8, 2014 at 15:06
  • $\begingroup$ $\tau_{yx}$ means torque about the $z$ axis and it is a horrible way of representing in since $\tau_{yx} = \tau_{xy} = \tau_{z}$ and torques are vectors and not tensors. Actually a torque is the position dependent component of the force screw. $\endgroup$ Aug 8, 2014 at 15:08

1 Answer 1


I think the lecture does not do a good job of explaining the fundamentals of mechanics. The (torque,force) combo works together the exact same way a (velocity,rotation) work together. These are sometimes called dual vectors, or screws.

So if a body is rotating, its velocity on the location you are measuring with $\vec{v} = \vec{r} \times \vec{\omega}$. The other way around is to think of linear velocity as describing the moment of rotation, by describing the center of rotation. In fact, given the two motion vectors $\vec{\omega}$ and $\vec{v}$ at a location A with position $\vec{r}_A$ the center of rotation is calculated at $$\vec{r}_C = \vec{r}_A + \frac{ \vec{\omega} \times \vec{v}}{|\vec{\omega}|^2}$$

Similarly with loading on a rigid body, the (equipollent) torque depends on location you are measuring it at with $\vec{\tau} = \vec{r} \times \vec{F}$. The other way around is to think of torque as the moment of force, by describing the center of force (also known as the line of action). In fact, given an applied force $\vec{F}$ and $\vec{\tau}$ at a location A with position $\vec{r}_A$ the line of action is located at $$\vec{r}_C = \vec{r}_A + \frac{ \vec{F} \times \vec{\tau} }{|\vec{F}|^2}$$

Now you see the duality between loading and motion, which is expressed in terms of the power (work rate) calculation $$\frac{{\rm d}}{{\rm d}t} W = \vec{v} \cdot \vec{F} + \vec{\omega} \cdot \vec{\tau} = (\vec{v},\vec{\omega}) \cdot (\vec{F},\vec{\tau}) $$

So velocity is the moment of rotation and torque to moment of force, both describing the location of a line in space. Together with their rotation and force vectors describing the direction of the line each forms a set of pluecker line coordinates.

In addition, to the line in space (like the rotation axis, or the line of force) they also provide information about the magnitude (amount of rotation $|\vec{\omega}|$, or force $|\vec{F}|$) and the pitch. The pitch in motion is the amount of parallel translation along the rotation axis that happens when the part is rotating (like a screw). The pitch in loading is the amount of parallel torque applied along the force axis when the part is loaded (also like a screw). The torque and velocity due to pitch is not position dependent like the amount due to the moment vectors. The pitch is found by

$$ \begin{align} h &= \frac{\vec{\omega}\cdot\vec{v}}{|\vec{\omega}|^2} & h &= \frac{\vec{F}\cdot\vec{\tau}}{|\vec{F}|^2}\end{align}$$

So if you find yourself confused about the components of torque, than try to imagine the dual of the situation with motion where $\vec{F} \rightarrow \vec{\omega}$ and $\vec{\tau} \rightarrow \vec{v}$ and see if it makes more intuitive sense. If the coordinates used to describe these geometric concepts are right handed, then the components are right handed also. Buy the math works either way, and if you are using a left hand system (I am looking at you Brits) the components will correspond accordingly.

As far as what index notation to apply, it is more of a personal choice than fundamentally describing something about the problem. Call it $\tau_z$ or $\tau_{xy}$ or the 3rd element of $\vec{\tau}$ or the 6th element of $(\vec{F},\vec{\tau})$ it is all the same.

To quote the link above

but in general you don't much care about such implementation details because the formulation of geometric algebra is coordinate free.


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